Find A⁻¹ and then X = A⁻¹B to solve three simultaneous equations.

Solve System of 3 Equations Using Matrix Inverse (AX=B)

Question

Solve the following system of equations using the matrix inverse method:

2x - 3y + 5z = 11

3x + 2y - 4z = -5

x + y - 2z = -3

This type appeared in CBSE 2025 Sample Paper and consistently shows up in board exam marking schemes with 5 marks allocated for the full working.

Solution — Step by Step

We express the system as a single matrix equation. Pull out the coefficient matrix $A$ , variable column $X$ , and constant column $B$ :

A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}

We must check that $|A| \neq 0$ before spending time on cofactors. Expanding along Row 1:

|A| = 2(2 \times -2 - (-4)(1)) - (-3)(3 \times -2 - (-4)(1)) + 5(3 \times 1 - 2 \times 1)

= 2(-4 + 4) + 3(-6 + 4) + 5(1) = 0 - 6 + 5 = -1

Since $|A| = -1 \neq 0$ , the inverse exists. We’re good to continue.

For each element $a_{ij}$ , the cofactor is $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the $2\times 2$ minor. Working through them row by row:

C_{11} = +(-4+4) = 0 \quad C_{12} = -(-6+4) = 2 \quad C_{13} = +(3-2) = 1

C_{21} = -(6-5) = -1 \quad C_{22} = +(-4-5) = -9 \quad C_{23} = -(2+3) = -5

C_{31} = +(12-10) = 2 \quad C_{32} = -(-8-15) = 23 \quad C_{33} = +(4+9) = 13

The adjugate (adjoint) is the transpose of the cofactor matrix — rows and columns swap:

\text{adj}(A) = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}

Now $A^{-1} = \dfrac{1}{|A|} \cdot \text{adj}(A) = \dfrac{1}{-1} \cdot \text{adj}(A)$ :

A^{-1} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}

X = A^{-1}B = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}

Row-by-row multiplication:

x = 0(11) + 1(-5) + (-2)(-3) = 0 - 5 + 6 = 1

y = -2(11) + 9(-5) + (-23)(-3) = -22 - 45 + 69 = 2

z = -1(11) + 5(-5) + (-13)(-3) = -11 - 25 + 39 = 3

Answer: $x = 1,\ y = 2,\ z = 3$

Always verify: $2(1) - 3(2) + 5(3) = 2 - 6 + 15 = 11$ ✓

Why This Works

The equation $AX = B$ is the matrix version of $ax = b$ . In scalars, we’d multiply both sides by $a^{-1}$ . Here, we premultiply both sides by $A^{-1}$ : left side gives $A^{-1}AX = IX = X$ , and right side gives $A^{-1}B$ . That’s the entire logic.

The determinant check at Step 2 is non-negotiable. If $|A| = 0$ , the matrix is singular — no unique solution exists, and the method breaks down entirely. Board examiners specifically award 1 mark for computing $|A|$ and stating $|A| \neq 0$ , so never skip it even if you “can see” the answer.

The adjugate transpose is where most errors happen. The cofactor matrix rows go down; the adjugate rows go across. Write the cofactors in a grid first, then transpose mentally — that habit prevents sign errors under exam pressure.

Alternative Method — Row Reduction (Augmented Matrix)

For those who find cofactors tedious, Gauss-Jordan elimination on the augmented matrix $[A | B]$ reaches the same answer faster in practice (though it’s not the method CBSE asks for in “matrix inverse” questions):

\left[\begin{array}{ccc|c} 2 & -3 & 5 & 11 \\ 3 & 2 & -4 & -5 \\ 1 & 1 & -2 & -3 \end{array}\right]

Apply $R_1 \leftrightarrow R_3$ , then eliminate to get row echelon form, then back-substitute. You’ll land at $x=1, y=2, z=3$ with less arithmetic. In JEE Main, where no method is mandated, this is often faster.

In CBSE boards, the question specifically says “using matrix inverse method” — use $A^{-1}B$ . Switching to row reduction will cost you method marks even if the answer is correct.

Common Mistake

Students write adj $(A)$ as the cofactor matrix directly, forgetting to transpose it. This gives wrong values for $A^{-1}$ and loses all 5 marks — correct method, wrong adjugate. Remember: cofactors fill the grid row by row, then you flip across the diagonal to get adj $(A)$ . The diagonal elements ( $C_{11}, C_{22}, C_{33}$ ) stay in place; everything else swaps. Double-check by verifying that $A \cdot A^{-1} = I$ whenever you have time.

Question

Solution — Step by Step

Why This Works

Alternative Method — Row Reduction (Augmented Matrix)

Common Mistake

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